1)A cyclist travels 40 km at a speed of xkm/h . Find the time taken in terms of x . Find the time taken when his speed is reduced by 2km/h . If the difference between the time is 1 hour , find the original speed x.
2)ABCD(rectangle corner) is a rectangle with AB =12cm(the length) and BC =7cm(the width) .AK=BL=CM=DN=x cm . If the area of KLMN is 54cm^2 find x.
有一個長方形用ABCD黎話4個角,咁長方形入面又有一個斜ge長方形用KLMN黎表示斜長方形ge 4個角.長方形ge長係12cm,闊係7cm.斜長方形面積係54cm^2 .AK,BL嗰D係兩個長方形相隔最少ge
(maths)Problem slove by quadratic equations
2006-10-19 5:42 am
回答 (2)
2006-10-19 5:52 pm
✔ 最佳答案
1)time taken to travel 40km at a speed of x km/h
= 40/x hrs
If speed becomes (x-2)km/h
time taken = 40/(x-2)
Time difference = 1
Therefore,
40/(x-2) - 40/x = 1
40x - 40(x-2) = x(x-2)
80 = x^2-2x
x^2-2x-80 = 0
(x-10)(x+8)=0
x=10 or x= -8
Since x is larger than 0, x=-8 is rejected
Therefore, the original speed = 10km/h
2)
Area of ABCD
= 12x7
= 84cm^2
Area of 4 Triangles AKN, BLK, CML, DNM
= Area of ABCD - Area of KLMN
Therefore,
x(7-x)/2 + x(12-x)/2 + x(7-2)/2 + x(12-x)/2 = 84-54
x(7-x) + x(12-x) = 30
19x-2x^2 = 30
2x^2-19x+30 = 0
(2x-15)(x-2)=0
x= 15/2 = 7.5 or x=2
Since x is less than 7, x= 7.5 is rejected
Therefore, x = 2
2006-10-19 5:59 am
1. time taken in terms of X
40/x hrs
40/x+1=40/(x-2)
(x-10)(x+8)=0
x=10 or -8(rejected)
x=10
original speed = 10km/h
40/x hrs
40/x+1=40/(x-2)
(x-10)(x+8)=0
x=10 or -8(rejected)
x=10
original speed = 10km/h
收錄日期: 2021-04-23 23:48:56
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